Let $f$ be a continuous and open map from $\mathbb R $ to $\mathbb R$.Prove that $f$ is monotonic.

Let $f$ be a continuous and open map from $\mathbb R $ to $\mathbb R$.Prove that $f$ is monotonic.

Suppose that $f$ is not monotonic.Then $\exists a,b;a>b$ such that $f(a)<f(b)$ and $c,d;c<d$ such that $f(c)>f(d)$.Since $f$ is continuous then there exists no break in the graph of $f$. But how should I use the fact that $f$ is open.Any help on how should I do the proof?

Solution 1:

Here is a direct proof:

$f$ has no local maximum or minimum

If $f$ has a local maximum at $x=a$, then there is $\delta >0$ such that $f(x)\le f(a)$ for all $x\in(a-\delta, a+\delta)$. Thus, $f$ maps the open interval $(x-\delta, x+\delta)$ to an interval of the form $(y,f(a)]$ or $[y,f(a)]$, neither of which is open. Alternatively, $f(a)$ is in the image of $(x-\delta, x+\delta)$, but no interval around $f(a)$ is totally contained in the image.

$f$ is injective

If $f(a)=f(b)$ with $a<b$, then $f$ is constant on $(a,b)$ and so $f$ maps $(a,b)$ to $\{f(a)\}$, which is not open.

$f$ is monotonic

Wlog, suppose $f(a)<f(b)$ with $a<b$. Take $c,d$ with $a<b<c<d$.

If $f(c)<f(a)$, then $f$ would also take the value $f(a)$ in $(b,c)$.

If $f(a)<f(c)<f(b)$, then $f$ would also take the value $f(c)$ in $(a,b)$.

Therefore, $f(b)<f(c)$.

Repeat the argument with $b,c,d$ and conclude that $f(c)<f(d)$.